A while back, my good friend Colin asked me to explain the red shift to him.
 
First off, I used the standard analogy of a speeding ambulance and how the pitch of the siren drops as it passes the observer.
 

 
Something along the lines of: The perceived pitch of the note is lower when the ambulance is moving away as the wave-fronts are reaching our ears less often, so as the frequency decreases, so does the wavelength (from λ = v/f). The opposite is true when the ambulance is approaching - the frequency of the wave reaching our ears is increased, hence the perceived wavelength increases too.
 
So far, so good.
 
I then moved on to the red shift and blue shift, explaining that a similar effect is occurring, albeit with light waves rather than sound waves. I even mentioned that the red shift wasn't actually discovered because receding galaxies appear red or redder (as in how would we know they aren't really that colour); rather, the effect was observed in spectrographs of receding objects.
 
Gaseous elements absorb light of a specific wavelength, and evidence of this absorption can be seen as dark "Fraunhofer" lines in the spectrum of stars and galaxies. Knowing enough about the composition of the gas around nearby (and therefore not receding) stars, these Fraunhofer lines form a 'signature' of the gases around a particular star.
 
What Hubble noticed were that in spectrographs of distant objects, these signature absorption lines did not appear at the corresponding wavelengths as observed in local objects (or observed in the lab). The lines appeared to be 'shifted' towards the red end of the spectrum (hence the name) and suggested the wavelengths absorbed were longer than they should be for some reason.
 

 
He postulated the effect was due to the Doppler effect, and inferred that these distant objects were receding from us.
 
I even showed him this video which illustrates the effect better than me:
 

 
Still so far, so good.
 
A week or so later, Colin came over and said there were a few aspects of all this that still didn't quite make sense, and could I further expound for him. And this is where I started floundering, hence why I need someone to clarify what I'm unable to explain.
 
His main concerns involved the speed of light, and how it must be observed the same for all observers. In other words, does the speed of the wave in question, be it a light wave or a sound wave, have a direct influence on the Doppler effect or red shift. What he was getting at I think, is why, if the speed of light remains constant, how is the wavelength 'stretched' as an object moves away. I know what he's getting at, but even knowing that the red shift does happen, I can't put into words how to account for his perceived discrepancy.
 
I waffled on about relativity, and why each wave front is reaching us slightly later because the object is moving away - hence the change in perceived wavelength. He seemed happy with that....
 
...until, a few days later, he posited something along these lines:
 
Take the diagram from the YouTube video above at around the 5:00 mark:
 

 
And here's the question I posted on his (our!) behalf:
 
====================
The finished diagram (around the 5:00 mark) is confusing as it actually shows the positions of the objects at sometime in the future; i.e. when the moving light source is further away from the observer. A more accurate representation should show the moving star at the same distance from the observer.

At this exact instant in time, when both objects are equidistant from the observer, and taking into account that the speed of light has to remain constant, the light emitted at that instant will reach the observer at the same time from both sources. Surely, at this instant in time, and if both objects are emitting the same frequency of light, the using λ = v/f shouldn't the wavelength remain the same? Hence the observer shouldn't see any red shift?
 ====================
 
Here's an amended diagram to illustrate my point:
 
 
 
 
I know I "Thanks in advance

Something along the lines of: The perceived pitch of the note is lower when the ambulance is moving away as the wave-fronts are reaching our ears less often, so as the frequency decreases, so does the wavelength (from λ = v/f). The opposite is true when the ambulance is approaching - the frequency of the wave reaching our ears is increased, hence the perceived wavelength increases too.

 
You have this wrong,  it should read..
 
Something along the lines of: The perceived pitch of the note is lower when the ambulance is moving away as the wave-fronts are reaching our ears less often, so as the frequency decreases, wavelength increases (from λ = v/f). The opposite is true when the ambulance is approaching - the frequency of the wave reaching our ears is increased, hence the perceived wavelength decreases.
Edit to include : 
To be honest, you're working it back to front. It should really read ...
Something along the lines of: The perceived pitch of the note is lower when the ambulance is moving away as the wave-fronts are reaching our ears less often, so as the wavelength increases, the frequency decreases (from λ = v/f). The opposite is true when the ambulance is approaching - the wavelength reaching our ears is decrease, hence the perceived pitch (frequency) increases.
------------------------------------------------------------------------
 
 
Given constant atmospheric conditions, the speed of sound through air is constant.  Imagine a parked car being passed by another car.  When the two cars are next to each other they both toot their horns.  You, the listener a distance away, would hear two different tones even if both car horns were producing the same note because relative to you, they are moving at different speeds.
 
Light works in exactly the same way.  Yes, the speed of light is constant in a particular medium. The published speed of light is as measured in a vacuum, ie, in space, but when travelling through glass or water or plastic or air it is different (slower)
 
Light is a bit of a strange animal, it can behave as either waves or particles or both at the same time. "red shift" is a demonstration of how it can behave like a wave.

I hope Kenny can mock this using the Coyote/Road Runner and "beep,beep" to explain....including charts.

The wavelength at any given frequency doesn't actually change... it is our ability to perceive it that changes.
 
let's say you are standing in knee high water at the beach and waves with a wavelength of 10 meters are peaking past you every 3 seconds. If you swim out vigorously through the waves they may begin to peak past you at 2 seconds but the 10 meter wavelength is still the same. Your perception of the frequency may be altered while the wavelength (and the speed of the wave's propagation) may remain constant.

mike_mccue
The wavelength at any given frequency doesn't actually change... it is our ability to perceive it that changes.
 
let's say you are standing in knee high water at the beach and waves with a wavelength of 10 meters are peaking past you every 3 seconds. If you swim out vigorously through the waves they may begin to peak past you at 2 seconds but the 10 meter wavelength is still the same. Your perception of the frequency may be altered while the wavelength (and the speed of the wave's propagation) may remain constant.

Sorry Mike, this is wrong.  You're mixing two difference points of reference.  If you measured the distance between wave peaks relative to you as you swam fast the length would appear shorter.

Yes, that is the basis of the theory of relativity.
 
As I explained, the swimmer's perception of the frequency would shift while out in reality the wavelength, and the speed of propagation, had not actually changed.
 
 

But "out in reality"  or  "as far as the swimmer is concerned" are two different view points which you can't mix.
 
The speed of a wave in water (the sea for instance) is determined by the depth of water.  Assuming the seabed is flat, the speed would be constant.  Therefore, given that the swimmer measures a shorter frequency and accepts the speed is constant the wavelength must be shorter as measured by the swimmer.
 
Someone in a boat traveling at a different speed in a different direction would get different results as well.
 
They are all "reality", and you can't pick and choose parts from different realities to prove anything.  Ask UB, he nose.

Reality proves nothing.
 
For that, nothing is very grateful.

The major difference is that light waves do not require a medium for travel, so the classical application of the Doppler effect doesn't apply precisely to this situation.
 
 

Relativistic Doppler Effect for Light

Consider two objects: the light source and the "listener" (or observer). Since light waves traveling in empty space have no medium, we analyze the Doppler effect for light in terms of the motion of the source relative to the listener.
We set up our coordinate system so that the positive direction is from the listener toward the source. So if the source is moving away from the listener, its velocity v is positive, but if it is moving toward the listener, then the v is negative. The listener, in this case, is always considered to be at rest (so v is really the total relative velocity between them). The speed of light c is always considered positive.
The listener receives a frequency f_L which would be different from the frequency transmitted by the source f_S. This is calculated with relativistic mechanics, by applying necessary the length contraction, and obtains the relationship:

f_L = sqrt [( c - v)/( c + v)] * f_S

Red Shift & Blue Shift

A light source moving
 
away from the listener ( v is positive) would provide an f_L that is less than f_S . In the visible light spectrum , this causes a shift toward the red end of the light spectrum, so it is called a red shift . When the light source is moving toward the listener ( v is negative), then f_L is greater than f_S . In the visible light spectrum , this causes a shift toward the high-frequency end of the light spectrum. For some reason, violet got the short end of the stick and such frequency shift is actually called a blue shift . Obviously, in the area of the electromagnetic spectrum outside of the visible light spectrum, these shifts might not actually be toward red and blue. If you're in the infrared, for example, you're ironically shifting away from red when you experience a "red shift."
 
 
I copied this...I'm shade blind.
 

Fraunhofer. Isn't that the same guy who invented MP3 encoding? Hardly anybody really understands that, either. I think I see a pattern here.